Problem: Here's a parameterization of a cylinder, for $0 < \theta < 2\pi$ : $\vec{v}(\theta, x) = (x, r\cos(\theta), r\sin(\theta))$ What is the inward-pointing vector normal to the area element of this cylinder given $r = 8$, $\theta = \dfrac{5\pi}{6}$, and $x = -2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\left( 0, -4\sqrt{3}, 4 \right)$ (Choice B) B $\left( -4\sqrt{3}, 4, 0 \right)$ (Choice C) C $\left( 0, 4\sqrt{3}, -4 \right)$ (Choice D) D $\left( 4\sqrt{3}, -4, 0 \right)$
Solution: The vector normal to the area element describes what's perpendicular to the surface, and the vector's magnitude represents the area of a tiny rectangle along the parameterization. After we compute it, we need to check whether it's pointing inwards or outwards. $\text{normal to area element} = \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial x}$ Let's calculate the cross product. $\begin{aligned} \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial x} &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ 0 & -r\sin(\theta) & r\cos(\theta) \\ \\ 1 & 0 & 0 \end{pmatrix} \\ \\ &= r\cos(\theta) \hat{\jmath} + r\sin(\theta) \hat{k} \end{aligned}$ Plugging in $r = 4$, $\theta = \dfrac{5\pi}{6}$, and $x = -2$, we get the vector normal to the area element: $\begin{aligned} \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial x} &= \left( 0, 8\cos\left(\dfrac{5\pi}{6}\right), 8\sin\left(\dfrac{5\pi}{6}\right) \right) \\ \\ &= \left( 0, -4\sqrt{3}, 4 \right) \end{aligned}$ The final step is to check whether this points inwards or outwards. The cylinder defined by $\vec{v}$ has an outward facing normal vector that points directly away from the origin, but with a zero $x$ -component. Because $\theta = \dfrac{5\pi}{6}$ corresponds to a zero $x$, negative $y$, and positive $z$, we should have a zero $x$, positive $y$, and negative $z$. We need to reverse the signs of our calculation to make it match the inward normal vector. Therefore, the inward-pointing vector normal to the area element is $\left( 0, 4\sqrt{3}, -4 \right)$.